Question

In this problem. we consider the delay introduced by the tcp slow-start phase. consider a client and a web server directly connected by one link of rate r. suppose the client wants to retrieve an object whose size is exactly equal to 15 s, where s is the maximum segment size (mss). denote the round-trip time between client and server as rtt (assumed to be constant). ignoring protocol headers. determine the time to retrieve the object (including tcp connection establishment) when
a. 4 sir> sir + rtt > 2sir
b. sir + rtt> 4 sir
c. sir> rtt.

Answer 1

The time that is required to retrieve the object (including TCP connection establishment) when $4S/R > S/R > RTT > 2S/R$ is equal to $\frac{14S}{R}+4RTT$

How to calculate the time needed to receive all objects.

In this scenario, we would denote the round-trip time that exist between the web server and the client with $RTT$ and assume that it is constant.

The time that is required to retrieve the object when $4S/R > S/R > RTT > 2S/R$ would be calculated by adding the round trip of sender and reciever together as follows:

Round trip = RTT + RTT

Next, we would determine the delay in packet transmission:

Note: $\frac{8S}{R}+\frac{4S}{R}=\frac{12S}{R}$

$Time = (\frac{S}{R} +RTT)+(\frac{S}{R} +RTT)+(\frac{12S}{R}) +RTT+RTT\\\\Time = \frac{14S}{R}+4RTT$

When $S/R +RTT > 4S/R$, we have:

The time that is required to retrieve the object would be calculated by adding the round trip of sender and reciever together as follows:

$Time = (\frac{S}{R} +RTT)+(\frac{S}{R} +RTT)+(\frac{8S}{R}) +RTT+RTT\\\\Time = \frac{11S}{R}+5RTT$

The time that is required to retrieve the object when $S/R > RTT$ would be calculated by adding the round trip of sender and reciever together as follows:

$Time = (\frac{S}{R} +RTT)+(\frac{14S}{R}) +RTT\\\\Time = \frac{15S}{R}+3RTT$

Read more on round trip here: brainly.com/question/6884622